Projectile:Launching and landing at different heights

1.If a shot is put an angle of 41 degrees relative to the horizontal with a velocity of 36 ft/s in the direction of the put, what will be the upward (vertical) velocity at the instant of release? What will be the forward (horizontal) velocity?
How high (above the point of release) will the shot go? What is the time it takes the shot to reach its maximum height?

2.If the shot in the problem above is released from a height of 6 ft and later lands on the ground (height = 0.0 ft),
what was the total time of flight? How far did the shot travel horizontally?

for the second part im not sure how to proceed do i use the following equation:
dy= viy^2 + 1/2at^2
6 = 23.6t -16t^t
olving for the quadratic equation i get two value for t
t1 = 1.14848205556934
t2 = 0.32651794443066

By common sense when the height was zero and if i was asked to calculate the total time it would have been 0.7375*2 = 1.475 s. So in part 2 the time should be greater than 1.475s since the height is included so can anyone put me on the correct track please ?

6 = 23.6t -16t^t
Here displacement and acceleration are in the same direction and initial velocity is in the opposite direction. So the equation becoms 6 = -23.6t + 16t^2. Now try

Can you explain to me how did you relate

Displacement , Acceleration and Initial velcoity together ? to me i wouldnt have solved it if it came up in a test. So what is the difference between part 2 and part 1 ; why did you take a = 32 and not -32 as i did in part 1.

t2= -b -sqrt(delta)2a
t2= (23.6-30.67)/(16*2)
t2 = -.0.22 it is not possible.
In the first part initial velocity and displacement are upwards and acceleration in the downwards. Out of three values mejority should be positive. In the second case displacement and acceleration are downwards and initial velocity is upwards. Hence the signs are applied.